Chap 3 Linear response theory

Ming-Che Chang

Department of Physics, National Taiwan Normal University, Taipei, Taiwan

(Dated: January 3, 2018)

I. GENERAL FORMULATION

A piece of matter would be polarized, or conduct cur-rent under an external electric field,

P = χeE, (1)

j = σE. (2)

If the electric field is not too strong, then the electric sus-ceptibility χe and the conductivity σ are independent ofthe electric field. They only depend on the material prop-erties in the absence of the electric field (i.e., in equilib-rium). This type of response is called the linear response.

The average of an observable (such as the electric po-larization) in equilibrium is

〈A〉0 =1

Z0

∑n

e−βE0n〈n0|A|n0〉. (3)

Under an external field, the states n0 are perturbed tobecome n, and the average becomes

〈A〉 =1

Z0

∑n

e−βE0n〈n|A|n〉 = 〈A〉0 + δ〈A〉. (4)

Our job is to find out δ〈A〉.Note that in the equation above, we do not alter the

thermal distribution. This is valid only if the perturba-tion is fast (compared to the time for reaching thermalequilibrium), such that the thermal distribution of thesystem are left unaltered. This is a tricky assumptionand the resulting response is called adiabatic response.On the other hand, if the perturbation is slow so thatthe system can be in equilibrium with the reservoir, thenthe response is called isothermal response. For in-depth discussions, see Sec. 3.2.11 of Ref. 1, and Sec. 8.3of Ref. 2.

In the following, as long as there is no ambiguity, wewill write the labels of a manybody state n simply asn. Before perturbation,

H0|n0〉 = E0n|n0〉, (5)

where E0n and |n0〉 are eigen-energies and eigenstates of

the manybody Hamiltonian H0. The external perturba-tion is assumed to be,

H(t) = H0 +H ′(t). (6)

After the perturbation (~ ≡ 1),

H(t)|n(t)〉 = i∂

∂t|n(t)〉. (7)

It is helpful to write,

|n(t)〉 = e−iH0t|nI(t)〉. (8)

Then,

H ′I(t)|nI(t)〉 = i∂

∂t|nI(t)〉. (9)

where

H ′I(t) ≡ eiH0tH ′e−iH0t, (10)

We say that the states and operators with subscript I arein the interaction picture.

To linear order,

|nI(t)〉 ' |n0I〉 − i

∫ t

−∞dt′H ′I(t

′)|n0I〉. (11)

Substitute it into Eq. (4), and keep only the terms tolinear order in H ′I , we have

〈A(t)〉

= 〈A〉0 − i∫ t

−∞dt′∑n

〈n0|[AI(t), H ′I(t′)]|n0〉e−βE0

n

Z0

= 〈A〉0 − i∫ t

−∞dt′〈[AI(t), H ′I(t′)]〉0, (12)

where

AI(t) ≡ eiH0tAe−iH0t. (13)

For example, if the following perturbation is turned onabruptly at t0,

H ′(t) =

∫dv B(r)︸︷︷︸

operator

· f(r, t)︸ ︷︷ ︸C−number

, (14)

then

δ〈A(r, t)〉 (15)

= −i∫dv′∫ t

t0

dt′〈[AI(r, t),BI(r′, t′)]〉0 · f(r′, t′)

= −i∫dv′∫ ∞t0

dt′θ(t− t′)〈[AI(r, t),BI(r′, t′)]〉0 · f(r′, t′).

This can be written as

δ〈A(x)〉 =

∫dx′∑α

χABα(x, x′)fα(x′), (16)

2

where x ≡ (r, t), dx′ ≡ dv′dt′, and

χABα(x, x′) = −iθ(t− t′)〈[AI(x), BIα(x′)]〉0. (17)

We will eventually let t0 → −∞, so that both spaceand time integrals cover the whole space-time. Eq (16)is called the Kubo formula, and χABα is called theresponse function. Be aware that the operators arewritten in the interaction picture.

II. DENSITY RESPONSE AND DIELECTRIC FUNCTION

A. Density response

In this section, we consider the perturbation of electrondensity caused by an external electric potential. Beforeperturbation,

H0 = T + VL + Vee, (18)

where VL is a one-body interaction, such as the electron-ion interaction, and Vee is the electron-electron interac-tion. The perturbation can be written in the followingform,

H ′ =

∫dvρe(r)φext(r, t), (19)

where ρe = −e∑s ψ†s(r)ψs(r) is the electron density, and

φext is an external potential.Because of the external potential, electron density

〈ρe〉0 → 〈ρe〉 = 〈ρe〉0 + δ〈ρe〉. (20)

Comparing with the Kubo formula, we find the followingreplacement,

A → ρe, (21)

B → ρe, (22)

f → φext. (23)

The Kubo formula gives

δ〈ρe(x)〉 =

∫dx′χρe(x, x

′)φext(x′), (24)

and the response function is

χρe(x, x′) = −iθ(t− t′)〈[ρe(x), ρe(x

′)]〉0. (25)

Remember that the operators are in the interaction pic-ture, but the subscript I is neglected from now on.

If the unperturbed system H0 is uniform in both spaceand time, then

χρe(x, x′) = χρe(x− x′). (26)

In this case, the convolution theorem in Fourier analysistells us that

δ〈ρe(κ)〉 = χρe(κ)φext(κ), (27)

where κ ≡ (q, ω), κx ≡ q · r− ωt, and

δ〈ρe(x)〉 =∑κ

eiκxδ〈ρe(κ)〉, (28)

δ〈ρe(κ)〉 =

∫dxe−iκxδ〈ρe(x)〉; (29)

φext(x) =∑κ

eiκxφext(κ), (30)

φext(κ) =

∫dxe−iκxφext(x). (31)

The summation over κ should be understood as∑κ

=1

V0

∑q

∫dω

2π. (32)

The Fourier expansion of the response function is

χρe(x− x′) ≡∑κ

eiκ(x−x′)χρe(κ), (33)

and

χρe(κ) =

∫d(x− x′)e−iκ(x−x′)χρe(x− x′)

= −i∫d(t− t′)θ(t− t′)eiω(t−t′) (34)

×∫d3(r− r′)e−iq·(r−r

′)〈[ρe(r, t), ρe(r′, t′)]〉0.

Since the system is uniform in space, one can perform anextra space integral 1

V0

∫d3r′ to the space integral above,

and use

1

V0

∫d3r′

∫d3(r− r′) =

1

V0

∫d3r

∫d3r′. (35)

Then it is not difficult to see that

χρe(κ) = − i

V0

∫ ∞0

dteiωt〈[ρe(q, t), ρe(−q, 0)]〉0. (36)

Note that ρe(−q, 0) can also be written as ρ†e(q, 0).In the following, we may sometimes use the particle

density ρ and its response function χρ, which are relatedto the electron density and its response function as

ρe = −eρ, χρe = e2χρ, (37)

and δ〈ρ〉 = −eχρφext. (38)

Also, note that these response functions are related to,but not the same as, the electric susceptibility χe men-tioned at the beginning of this chapter.

B. Dielectric function

The response function connects δρe with φext. How-ever, the dielectric function connects φext with the to-tal potential φ, which is the sum of φext and the potentialdue to material response,

φ(κ) = φext(κ) + δφ(κ), (39)

3

and

ε(κ) =φext(κ)

φ(κ). (40)

The total particle density is

〈ρ〉 = 〈ρ〉ext + δ〈ρ〉. (41)

Particle densities are related to the potentials via thePoisson equations,

q2φext(κ) = −4πe〈ρ(κ)〉ext, (42)

q2φ(κ) = −4πe〈ρ(κ)〉. (43)

Note that quantities such as φext(κ) = φext(q, ω) is al-lowed to be frequency-dependent. If SI is used, then justreplace 4π with 1

ε0.

Combine the three equations above, we get a relationbetween φ and φext,

φ(κ) = φext + 4πe2χρφextq2

. (44)

This leads to a relation between ε and χρ,

1

ε(κ)= 1 +

4πe2

q2︸ ︷︷ ︸V (2)(q)

χρ, (45)

in which V (2)(q) is the Fourier transform of the Coulombpotential energy, V (2)(r) = e2/r.

Instead of using δ〈ρ〉 = −eχρφext, an alternative rela-tion is,

δ〈ρ〉 = −eχ0ρφ, φ = φext + δφ. (46)

It’s not difficult to see that,

χρ =χ0ρ

1− 4πe2

q2 χ0ρ

, (47)

and

ε(κ) = 1− 4πe2

q2χ0ρ. (48)

Thus, given φext = 4πe2/q2 ≡ V (q), one has

Veff (q) =V (q)

ε(q)=

V (q)

1− V (q)χ0ρ(q)

. (49)

The calculation of χρ is based on Eq. (36), in whichone averages over unperturbed manybody states (includ-ing electron interactions). A great advantage of usingthe alternative response function χ0

ρ is that, since the lo-cal field correction has been included in φ, one mayuse non-interacting manybody states in the calculation ofthe response function. This is justified as follows (Ref. 3):

The interaction term is, apart from a one-body correc-tion (see Sec. IV.B.1 of Chap 1),

Vee =1

2

∫dvdv′V (2)(r− r′)ρe(r)ρe(r

′). (50)

Use the mean field approximation, and expand the chargedensity with respect to a mean value 〈ρ(r)〉e,

ρe(r) = 〈ρe(r)〉+ ρe(r)− 〈ρe(r)〉︸ ︷︷ ︸δρe(r)

. (51)

Neglecting the (δρe)2 term, we have

Vee '∫dvdv′V (2)(r− r′)ρe(r)〈ρe(r′)〉 (52)

− 1

2

∫dvdv′V (2)(r− r′)〈ρe(r)〉〈ρe(r′)〉.

The mean-field Hamiltonian under perturbation is(dropping the second term in Eq. (52)),

HMF

= H0 +

∫dvdv′V (2)(r− r′)ρe(r)〈ρe(r′)〉+

∫dvρe(r)φext

= H0 +

∫dvρe(r)φ(r), (53)

where H0 = T + VL, and

φ(r) = φext(r) +

∫dv′V (2)(r− r′)〈ρe(r′)〉. (54)

The second term in φ(r) is the induced potential, or thelocal field correction. That is, if one calculates the re-sponse to the total perturbing potential φ(r), then the

unperturbed system is H0, which is non-interacting.

C. Calculation of χ0ρ

We now drop the superscript and subscript 0 that referto equilibrium states. Recall that

χ0ρ(κ) = − i

V0

∫ ∞0

dteiωt〈[ρ(q, t), ρ(−q, 0)]〉. (55)

In the interaction picture, ρ(q, t) = eiH0tρ(q)e−iH0t. Thesummation

I(q; t, 0)

≡∑n

e−βEn

Z〈n|ρ(q, t)ρ(−q, 0)|n〉 (56)

=∑n,m

e−βEn

Zei(En−Em)t〈n|ρ(q, 0)|m〉〈m|ρ(−q, 0)|n〉,

where we have inserted a complete set∑m |m〉〈m|, and

used e−iH0t|m〉 = e−iEmt|m〉.

4

Since both |n〉 and |m〉 are manybody states of non-

interacting particles, 〈m|a†ksak−q,s|n〉 can be non-zeroonly if, when comparing with |n〉, the |m〉 state has onemore electron at state (k, s), but one less electron at(k − q, s). Therefore, En − Em = −εk + εk−q, a dif-ference of two single-particle energies. It follows that,

I(q; t, 0) (57)

=∑n

e−βEn

Z

∑k,s

ei(εk−q−εk)t〈n|a†k−q,saksa†ksak−q,s|n〉.

For free-particle or Hartree-Fock-like states (consider q 6=0),

〈a†k−q,saksa†ksak−q,s〉 = 〈a†k−q,sak−q,s〉〈aksa

†ks〉. (58)

This is related to the Wick theorem at finite tempera-ture. See, e.g., Appendix 3 of Ref. 1 and related discus-sion. The q = 0 term, if not dropped, would be cancelledout in Eq. (63) below anyway. Now, it is left as an exer-cise to show that,

f(εk) ≡ 〈a†ksaks〉 =∑n

e−βEn

Z〈n|a†ksaks|n〉

=1

1 + eβεk. (59)

Thus,

I(q; t, 0) = 2∑k

ei(εk−q−εk)tf(εk−q)[1− f(εk)]. (60)

Had grand canonical ensemble been used, we’d have

f(εk) =1

1 + eβ(εk−µ). (61)

This is the Fermi-Dirac distribution function (spin-independent here).

Similarly, one can show that,

I(−q; 0, t) = 2∑k

ei(εk−q−εk)tf(εk)[1− f(εk−q)]. (62)

From Eq. (55), we have

χ0ρ(κ) = − i

V0

∫ ∞0

dteiωt[I(q; t, 0)− I(−q; 0, t)] (63)

= − 2i

V0

∑k

∫ ∞0

dteiωtei(εk−q−εk)t[f(εk−q)− f(εk)].

The integral over time is∫ ∞0

dtei(ω+iδ)tei(εk−q−εk)t =i

ω + iδ + (εk−q − εk).

(64)The positive infinitesimal δ is added to ensure the con-vergence of the exponential at t =∞.

FIG. 1 Electric susceptibility of electron gas in different di-mensions. Fig from Dugdale’s Phys Scr, 2016.

Finally,

χ0ρ(q, ω) =

2

V0

∑k

f(εk−q)− f(εk)

ω + iδ + (εk−q − εk), (65)

and

ε(q, ω) = 1− 4πe2

q2

2

V0

∑k

f(εk−q)− f(εk)

ω + iδ + (εk−q − εk). (66)

This is called the Lindhard dielectric function.

1. Low-frequency limit

For frequency as low as ω vF q, the ω in the denom-inator can be neglected, and

χ0ρ(q, 0) ' 2

V0

∑k

f(εk−q)− f(εk)

εk−q − εk. (67)

Recall that we are calculating the adiabatic response(see Sec. I), which presumes the perturbation to be fast.Thus, it seems risky to apply our result to the low-frequency limit. However, for the density response, theresult reported below is indeed valid. See p. 136 of Ref. 1for an explanation.

At long wavelength,

χ0ρ(q, 0) ' − 2

V0

∑k

(−∂fk∂ε

)= −D(εF ), (68)

where D(εF ) is the density of states at the Fermi energy.For 3D free electron gas,

D(εF ) =mkFπ2~2

. (69)

In this limit, the dielectric function is

ε(q, 0) = 1 +k2TF

q2, (70)

5

where k2TF = 4πe2D(εF ) is the Thomas-Fermi wave

vector. For Copper, kTF ' 1.8 × 1010/m. Thus thescreening length 1/kTF ' 0.55A.

Given a point charge with φext(r) = Q/r, the screenedelectrostatic potential is,

φ(r) =

∫d3q

(2π)3

φext(q)

ε(q, 0)eiq·r

=

∫d3q

(2π)3

4πQ

q2 + k2TF

eiq·r

=Q

re−kTF r. (71)

For general wavelength (in 3D), it can be shown that(see Sec. 14 of Ref. 4),

χ0ρ(q, 0) ' −D(εF )F

(q

2kF

), (72)

where

F (x) =1

2+

1− x2

4xln

∣∣∣∣1 + x

1− x

∣∣∣∣ (73)

is the Lindhard function (see Sec. III.A). The slope ofF (q/2kF ) has a logarithmic singularity at q = 2kF (seeFig. 1), resulting in a sudden decrease of screening. Thisis related the fact that the pair creation energy, δεq =

εk+q − εk = ~2

m q(±kF + q/2), can no longer be zero onceq > 2kF (see Fig. 3(a)).

It can be shown that (not easy), if we use Eq. (72), thenfor kF r 1, the induced charge density is (see Fig. 2),

δρ(r) ' cos(2kF r)

r3. (74)

One can see p. 178 of Ref. 4 for a derivation. The os-cillation of δρ (and accompanied potential variation) iscalled the Friedel oscillation.

For reference, we show the electric susceptibility forelectron gas in lower dimensions (see Fig. 1). In 2D, itis,

χ0ρ(q) =

− mπ~2 if q ≤ 2kF ,

− mπ~2

[1−

√1− (2kF /q)2

]if q > 2kF .

(75)In 1D, it is,

χ0ρ(q) = − 2m

π~2

1

qln

(q + 2kFq − 2kF

), (76)

which diverges at q = 2kF . One can find more details in,e.g., Sec. 4.4 of Ref. 1.

2. High-frequency limit

The response function in Eq. (65) can be rewritten as,

χ0ρ(q, ω) =

2

V0

∑k

f(εk)2(εk−q − εk)

ω2 − (εk−q − εk)2. (77)

FIG. 2 Friedel oscillation. Fig from Chazalviel’s Coulombscreening by mobile charges.

(a) (b)

FIG. 3 (a) Predicted dispersion curve of the plasmon (insodium) vs electron-hole continuum. (b) Measured plasmondispersion in sodium. The vertical bars are measured widthsof the plasmon resonances. Figs are from Ref. 5.

For high frequency and long wavelength (ω vF q),

χ0ρ(0, ω) ' q2

mω2

2

V0

∑k

f(εk) =q2n

mω2, (78)

where n is the particle density. Therefore,

ε(0, ω) = 1−ω2p

ω2, (79)

where ω2p = 4πne2/m is the plasma frequency. For

copper, n = 8 × 1022/cm3, and the plasma frequency isωp = 1.6× 1016/s (corresponding to a wavelength about

1200 A). A piece of metal becomes transparent to an EMwave with frequency ω if ω > ωp.

It is possible to consider the shift of plasma frequencywhen q is finite. To leading order, we have

ε(q, ω) ' 1− 4πe2

mω2

2

V0

∑k

f(εk)

(1 +

~2q2

m2ω2k2

). (80)

At T = 0,

2

V0

∑k

f(εk)k2 =3

5nk2

F . (81)

It follows that,

ε(q, ω) ' 1−ω2p

ω2

[1 +

3

5

(~kFmω

)2

q2

]. (82)

6

There is longitudinal plasma oscillation when ε(q, ω) = 0.Thus,

ω2 = ω2p +

3

5v2F q

2 + · · · . (83)

See Fig. 3 for details. The electron-hole continuum inFig. 3(a) is the domain where the creation of an electron-hole pair out of a Fermi sphere is allowed (more in Chap6).

For reference, the plasma frequency for 2D electron gasis,

ω2 =2πne2

mq +

3

4v2F q

2 + · · · , (84)

which is gapless. See p. 204 of Ref. 1 for more details.Finally, note that (see Eqs. (70), (79))

limq→0

limω→0

ε(q, ω) 6= limω→0

limq→0

ε(q, ω). (85)

That is, the dielectric function is not analytic at (q, ω) =(0, 0).

III. CURRENT RESPONSE AND CONDUCTIVITY

A. Current response

In this section, we consider the electric current drivenby an external electric field. Before perturbation,

H0 =

∫dvψ†(r)

p2

2mψ(r) + VL + Vee, (86)

where VL is the one-body potential energy, and Vee is theelectron interaction. In general, the external electric fielddepends on both scalar and vector potentials,

E(r, t) = −∇φ− 1

c

∂A

∂t. (87)

For a static field, it is common to use E = −∇φ, as inEq. (19). A static and uniform field then has φ(r) =−E · r, A(r) = 0. A disadvantage of this scalar potentialis that it is not bounded at infinity. To avoid such aproblem, one can choose a gauge such that

E(r, t) = −1

c

∂A

∂t. (88)

In this case, a static and uniform field has A(t) = −cEt,and φ(r) = 0.

After applying the electric field, the Hamiltonian be-comes,

H =

∫dvψ†(r)

(p + e

cA)2

2mψ(r) + VL + Vee

= H0 +e

2mc

∫dv(ψ†p ·Aψ + ψ†A · pψ

)+

e2

2mc2

∫dvA2ψ†ψ, (89)

where H0 refers to the parts that do not depend on A.The particle current density operator J is related to thevariation of the Hamiltonian as follows,

δH =e

c

∫dvJ · δA, (90)

where

J ≡ 1

2mi

[ψ†∇ψ −

(∇ψ†

)ψ]

︸ ︷︷ ︸paramagnetic current Jp

+e

mcAψ†ψ.︸ ︷︷ ︸

diamagnetic current JA

(91)We would like to find out the connection between 〈J〉

and A (to first order). After perturbation, a manybodystate

|n0〉 → |n〉 ' |n0〉+ |n1〉, (92)

where |n1〉 is of order A. Therefore,

〈n|J|n〉 = 〈n0|JA|n0〉+ 〈n0|Jp|n1〉+ 〈n1|Jp|n0〉+O(A2).(93)

We have assumed, of course, that the equilibrium statecarries no current, 〈n0|Jp|n0〉 = 0.

After taking the thermal average, the first term be-comes,

〈JA〉 =e

mcA(r, t)ρ(r). (94)

The other two terms are evaluated using the Kubo for-mula in Eq. (16), with the following replacement,

A → Jpα, (95)

B → Jp, (96)

f → e

cA. (97)

This gives us (recall that x = (r, t))

〈Jpα(x)〉 =e

c

∫dx′χpαβ(x, x′)Aβ(x′), (98)

where

χpαβ(x, x′) = −iθ(t− t′)⟨

[Jpα(x), Jpβ(x′)]⟩. (99)

After combining with the diamagnetic term in Eq. (94),the response function for the total current is,

χαβ(x, x′) = δαβδ(x− x′)ρ(x)

m+ χpαβ(x, x′). (100)

Since H0 is time independent, ρ(x) = ρ(r), and theresponse function χαβ(x, x′) = χαβ(r, r′; t − t′). Apply-ing the convolution theorem to the time variable (seeEq. (27)), one has

〈Jα(r, ω)〉 =e

c

∫dv′χαβ(r, r′;ω)Aβ(r′, ω), (101)

7

where

χαβ(r, r′;ω) = δαβδ(r− r′)ρ(r)

m+ χpαβ(r, r′;ω), (102)

in which

χpαβ(r, r′;ω) = −i∫dtθ(t)eiωt〈[Jα(r, t), Jβ(r′, 0)]〉.

(103)The vector potential is related to the electric field as fol-lows,

E(ω) = iω

cA(ω). (104)

Therefore, for the electric current density Je = −eJ, onehas

〈Jeα(r, ω)〉 =

∫dv′σαβ(r, r′;ω)Eβ(r′, ω). (105)

The conductivity tensor is

σαβ(r, r′;ω) = ie2

ωχαβ(r, r′;ω). (106)

Since the conductivity in general is a non-local quantity,the current density at point r would not only depend onthe electric field at r, but also on neighboring electricfield.

For a homogeneous material,

σαβ(r, r′;ω) = σαβ(r− r′;ω). (107)

We can then apply the convolution theorem to the spacevariable and get

〈Jeα(q, ω)〉 = σαβ(q, ω)Eβ(q, ω), (108)

where

σαβ(q, ω) = ie2

ω

[δαβ

ρ0

m+ χpαβ(q, ω)

], (109)

in which ρ0 is the homogeneous particle density, and (Cf.Eq. (36))

χpαβ(q, ω) = − i

V0

∫dtθ(t)eiωt〈[Jα(q, t), Jβ(−q, 0)]〉.

(110)Note that the diamagnetic part diverges as ω → 0. Forusual conductors and insulators, this divergence wouldbe cancelled by part of the paramagnetic term, so thatthe DC conductivity remains finite. In a superconduc-tor, which is a perfect diamagnet (think of the Meissnereffect), the paramagnetic term vanishes in the DC limit,and the conductivity is purely imaginary,

σSCαβ (q, ω) = ie2

ωδαβ

ρ0

m. (111)

A purely imaginary conductivity leads to inductive be-havior, and would not cause energy dissipation.

B. Electrical conductivity

We would like to start from a formulation that doesnot presume spacial homogeneity:

χpαβ(r, r′;ω) = −i∫ ∞

0

dteiωt⟨[Jpα(r, t), Jpβ(r′, 0)

]⟩,

(112)where Jpα(r, t) = eiH0tJpα(r)e−iH0t. Therefore,

Iαβ(r, t, r′, 0)

≡∑n

e−βEn

Z〈n|Jpα(r, t)Jpβ(r′, 0)|n〉 (113)

=∑n,m

e−βEn

Zei(En−Em)t〈n|Jpα(r, 0)|m〉〈m|Jpβ(r′, 0)|n〉.

We have inserted a complete set∑m |m〉〈m|, and used

e−iH0t|m〉 = e−iEmt|m〉.The particle current density operator can be written

as (see Chap 1),

Jpα(r) =∑µν

〈µ|J (1)α (r)|ν〉a†µaν , (114)

where J(1)α (r) is the one-body operator that can be found

in the Table of Chap 1.

From now on, assume the electrons are not interactingwith each other. Substitute Jpα(r) into Eq. (113), we getterms with the form,

〈n|a†1a2|m〉〈m|a†3a4|n〉, (115)

where 1, 2 · · · are simplified notations for single-particlestate labels µ, ν. For this type of term to be non-zero,the single-particle states have to satisfy (1 = 4, 2 = 3),or (1 = 2, 3 = 4). They both lead to En − Em = ε1 − ε2

(the second case has ε1 = ε2).

The summation over m can now be removed, and

Iαβ(r, t, r′, 0) =∑

1,2,3,4

ei(ε1−ε2)t〈1|J (1)α |2〉〈3|J

(1)β |4〉

× 〈a†1a2a†3a4〉(δ14δ23 + δ12δ34). (116)

The thermal averages are (see Eq. (59)),

〈a†1a2a†2a1〉 = f1(1− f2), (117)

〈a†1a1a†2a2〉 = f1f2. (118)

where f is the Fermi-Dirac distribution function. As aresult, one can show that

Iαβ(r, t, r′, 0)− Iβα(r′, 0, r, t)

=∑12

ei(ε1−ε2)t〈1|J (1)α |2〉〈2|J

(1)β |1〉(f1 − f2). (119)

8

FIG. 4 Real part of the optical conductivity for alkali metals.The Drude peaks can be seen at low frequency. Smith, PhysRev B 2, 2840 (1970).

Therefore,

χPαβ(r, r′, ω)

= −i∫ ∞

0

dteiωt[Iαβ(r, t, r′, 0)− Iβα(r′, 0, r, t)]

=∑12

(f1 − f2)〈1|J (1)

α (r)|2〉〈2|J (1)β (r′)|1〉

Ω + ε1 − ε2, (120)

where Ω ≡ ω + iδ.If the material is homogeneous, then

χPαβ(q, ω) =1

V0

∑12

(f1 − f2)〈1|J (1)

α (q)|2〉〈2|J (1)β (−q)|1〉

Ω + ε1 − ε2,

(121)in which (see Chap 1),

J (1)α (q) =

1

2m

(pαe−iq·r + e−iq·rpα

). (122)

1. Uniform limit

For the uniform case (q = 0), the conductivity is(rewrite 1, 2 as µ, ν),

σαβ(0, ω) (123)

=ie2

ω

[δαβ

ρ0

m+

1

m2V0

∑µν

(fµ − fν)〈µ|pα|ν〉〈ν|pβ |µ〉

Ω + εµ − εν

].

The denominator can be decomposed as

1

εµν ± Ω=

1

εµν

(1∓ Ω

εµν ± Ω

), (124)

where εµν ≡ εµ − εν . Substitute this to Eq. (123), thenthe first term of the decomposition would cancel with thediamagnetic term, because of the following f -sum rule:

1

V0

∑µν

(fµ − fν)〈µ|pα|ν〉〈ν|pβ |µ〉

εµν= −mρ0δαβ . (125)

As a result,

σαβ(0, ω) =e2

im2V0

∑µν

(fµ−fν)〈µ|pα|ν〉〈ν|pβ |µ〉εµν(Ω + εµν)

. (126)

Let’s apply this to a metal, and consider only the intra-band contribution, such that |µ〉, |ν〉 are the Bloch states|nk〉, |nk′〉 (see p. 414 of Ref. 6). For small q, but beforetaking q = 0, we have

〈nk′|eiq·rpβ |nk〉 ' mvβδk′,k+q. (127)

Thus,

σαβ(q, ω) ' 2e2

iV0

∑k

fk − fk+q

ωk − ωk+q

vαvβωk − ωk+q + ω + iδ

.

(128)For q k, we have

fk+q − fkωk+q − ωk

' ∂f

∂ε. (129)

It follows that,

σαβ(q, ω)

' 2ie2

∫d3k

(2π)3

(−∂f∂ε

)vαvβ

ωk − ωk+q + ω + iδ

= 2e2

∫d3k

(2π)3

(−∂f∂ε

)vαvβτ

1− iτ(ω − q · v), (130)

in which we have used ωk − ωk+q ' −q · v, and therelaxation time τ = 1/δ. This agrees with the resultbased on the Boltzmann equation (e.g., see Sec. 23.2 ofMarder’s).

Finally, in the uniform limit,

σαβ(0, ω) =2ie2

ω + iτ−1

∫d3k

(2π)3

(−∂f∂ε

)vαvβ . (131)

When ωτ 1, the real part of the conductivity ap-proaches a Dirac delta function. Define a broadened deltafunction as,

δτ (ω) =1

π

τ

1 + (ωτ)2, (132)

then,

Re σαβ(0, ω) = D δτ (ω) + Re σRegαβ (0, ω). (133)

The coefficient D is known as as the Drude weight (seeFig. 4). We have added a possible regular part at higherfrequency due to inter-band transition.

9

FIG. 5 Quantized Hall conductivity of a 2D electron gas instrong magnetic field.

Remark: For the longitudinal conductivity, in general,

limq→0

limω→0

σαα(q, ω) 6= limω→0

limq→0

σαα(q, ω). (134)

To get the correct DC conductivity, one needs to take thelimit on the RHS. For the limit on the left, the energylevels are discrete while the system is still finite (finiteq). As a result, inter-level transitions would be frozen inthe DC limit (ω → 0), and the current vanishes.

2. Hall conductivity, uniform and static

Finally, we would like to consider the DC Hall con-ductivity. According to Eq. (126), it can be re-writtenas

σDCα6=β =e2

iV0

∑µν,µ 6=ν

fµ〈µ|vα|ν〉〈ν|vβ |µ〉 − 〈µ|vβ |ν〉〈ν|vα|µ〉

ε2µν

.

(135)If the single-particle states are Bloch states (µ → nk),〈r|nk〉 = eik·runk(r), where unk(r) is the cell-periodic

function. Define Hk ≡ e−ik·rH(r,p)eik·r, then

〈nk| pm|n′k〉 = 〈unk|

p + ~k

m|un′k〉

= 〈unk|∂Hk

~∂k|un′k〉. (136)

With the help of the identity (for µ 6= ν),

〈unk|∂Hk

∂k|un′k〉 = (εnk − εn′k) 〈∂unk

∂k|un′k〉, (137)

one can show that,

σDCα6=β =e2

~V0

∑nk

fnk1

i

(⟨∂unk∂kα

|∂unk∂kβ

⟩−⟨∂unk∂kβ

|∂unk∂kα

⟩)︸ ︷︷ ︸

Berry curvature Ωγnk

,

(138)

where α, β, and γ are cyclic. This is also known as theTKNN formula (see Ref. 7).

In 2D, for a filled band (usually a Landau subband),

C(n)1 ≡ 1

2π

∫filled BZ

d2kΩznk (139)

must be an integer (see Sec. II.B of Ref. 8). Therefore,the Hall conductivity from filled bands is quantized,

σDCxy =e2

h

∑filled n

C(n)1 . (140)

The quantized (topological) nature of such an inte-gral is first shown by D.J. Thouless to explain theinteger quantum Hall effect (see Fig. 5). Notethat h/e2 ' 25.813 kΩ, and quantized Hall resistanceRH = h

e2 /i, where i is a positive integer.

1. (a) Show that the Fourier transformations ofthe Coulomb potential φ(r) = 1/r in 3D and 2D are,

3D : φ(q) =4π

q2; 2D : φ(q) =

2π

q. (141)

(b) Show that the density of states of free electron gasesin 3D and 2D are,

3D : D(εF ) =mkFπ2~2

; 2D : D(εF ) =m

π~2. (142)

2. Put a point charge Q in a 2D electron gas.(a) Under the long-wavelength (Thomas-Fermi) approxi-mation (i.e., χρ(q) = −m/π~2 in Eq. (75) is used), showthat the screened electrostatic potential at kF r 1 is,

φ(r) =Q

r−Qπq0

2[H0(q0r)−N0(q0r)], (143)

where q0 = 4πme2/h, H0(x) is a Struve function, andN0(x) is a von Neumann function. Related integral canbe found in, for example, Gradshteyn and Ryzhik’s Tableof integrals, series, and products.(b) Following (a), show that at large distance,

φ(r) ' Q

q20r

3. (144)

Note: If the complete form of Eq. (75) is used, then

φ(r) ' Q sin(2kF r)

(2kF r)2, (145)

which shows the Friedel oscillation. Ref: Frank Stern,Phys. Rev. Lett. 18, 546 (1967). Also, p. 225 of Ref. 1.3. Assume H0 = p2/2m+ VL(r). With the help of

[H0, rα] =~im

pα, (146)

10

derive the f -sum rule,

1

V0

∑µν

(fµ − fν)〈µ|pα|ν〉〈ν|pβ |µ〉

εµν= −mρ0δαβ , (147)

in which ρ0 is the particle density.4. Start from Eq. (126), derive the conductivity sum rule,∫ ∞

−∞Re σαα(ω)dω =

1

4ω2p. (148)

That is, the area below a curve in Fig. 4 is fixed by theplasmon frequency. More sum rules can be found in, e.g.,Chap 4 of Pines and Nozieres, The theory of quantumliquids, Addison-Wesley Publishing, 1989.

References

[1] G.F. Giuliani and G. Vignale, Quantum theory of the elec-tron liquid, Cambridge University Press, 2005.

[2] L.P. Levy, Magnetism and superconductivity, Springer-Verlag, 2000.

[3] Sec. I.5 of T. Giamarchi, A. Iucci, and C. Berthod, Intro-duction to Many body physics, on-line lecture notes.

[4] A.L. Fetter, J.D. Walecka, Quantum Theory of Many-Particle Systems, Dover Books, 2003.

[5] Chap 6: Electron Transport, by P. Allen, in ConceptualFoundations of Materials: A standard model for ground-and excited-state properties, ed. by S.G. Louie and K.L.Choen, Elsevier Science 2006.

[6] G. Grosso and G.P. Parravicini, Solid State Physics, Aca-demic Press, 2000.

[7] D.J. Thouless, M. Kohmoto, P. Nightingale, and M. deNijs, Phys. Rev. Lett. 49, 405 (1982).

[8] D. Xiao, M.C. Chang, and Q. Niu, Rev. Mod. Phys. 82,1959 (2010).